Partitions, Darboux sums and integrals

Try

For f(x)=x on [0,1], divide the interval into n equal pieces. On each piece the left endpoint gives the infimum and the right endpoint the supremum. Lower and upper rectangle sums trap any proposed area. Their difference is 1/n, so the trap can be made arbitrarily narrow. This gives an integral definition based on order and completeness, before introducing any antiderivative.

Predict

Predict what happens when an additional partition point is inserted. A smaller interval has fewer function values, so its infimum cannot be lower and its supremum cannot be higher than those of the original interval. The lower sum can only rise, while the upper sum can only fall. Refinement improves enclosure even if no formula for an extremum is available.

Experiment

Compute equal-partition bounds.

Number of pieces n	Lower sum	Upper sum	Gap
1	0	1	1
2	1/4	3/4	1/2
4	3/8	5/8	1/4

The general lower sum is (n−1)/(2n) and upper sum (n+1)/(2n), both approaching 1/2. This argument establishes the integral of x from sums. A graph with finitely many rectangles illustrates the enclosure, while the formula for arbitrary n proves the gap can be made as small as requested.

Understand

A partition P is a finite ordered list a=x₀<x₁<⋯<x_n=b. For bounded f, let m_i and M_i be its infimum and supremum on the ith closed subinterval. Define L(f,P)=Σm_iΔx_i and U(f,P)=ΣM_iΔx_i. Lower and upper integrals are the supremum of all lower sums and infimum of all upper sums. Boundedness guarantees these quantities are finite.

For any two partitions P,Q, their union is a common refinement R. Monotonicity gives L(P)≤L(R)≤U(R)≤U(Q). Hence the lower integral never exceeds the upper integral. Darboux integrability means equality. The common value is the integral; it has not been defined as a primitive difference.

Apply

Prove the useful criterion: f is integrable exactly when every ε>0 permits a partition with U−L<ε. If such partitions exist, the difference between upper and lower integrals is nonnegative and smaller than every positive ε, so it is zero. Conversely, if both equal I, choose a lower sum above I−ε/2 and an upper sum below I+ε/2. A common refinement combines those approximations into a single small-gap partition.

Repair “every bounded function is integrable because every partition has finite sums.” The indicator of Q on [0,1] has infimum zero and supremum one on every nondegenerate interval by density. Thus every lower sum is zero and every upper sum one, leaving a permanent gap.

From an integral average to a certified finite sum

For f(x)=x² on [0,1], the average value is 1/3. It equals f(c) at c=√(1/3), but that point need not coincide with a chosen partition tag. A finite sum uses specified sample locations; the mean-value theorem for integrals asserts the existence of some representative value. These are different claims.

Compare the measured error with the proved bound

Function x²: curved continuous function2x+1: sharp endpoint exampleStep at 1/2: missing hypothesis Partition Uniform xⱼ=j/nGraded xⱼ=(j/n)² Cells 24816 Tag location LeftMidpointRight

Predict the error sign before changing tags. The rectangle area uses the actual width of each cell; graded cells are not equally weighted. Variables are dimensionless.

Rectangle sum Q=0.46875000; exact integral I=0.33333333; signed error Q−I=0.13541667. Maximum cell width: 0.25.

Proved absolute truncation bound: 0.25000000, using the common slope bound M=2. This bound need not equal the observed error; floating-point rounding is separate.

The average I/(1−0) is attained at c≈0.57735027. This follows from continuity and the intermediate value theorem, not from choosing a particular rectangle tag.

Cell	Tag	Contribution	Exact cell integral	Signed error	Local bound
0 to 0.25	0.25	0.015625	0.005208	0.010417	0.062500
0.25 to 0.5	0.5	0.062500	0.036458	0.026042	0.062500
0.5 to 0.75	0.75	0.140625	0.098958	0.041667	0.062500
0.75 to 1	1	0.250000	0.192708	0.057292	0.062500

Reset

At most sixteen cells and sixty-five curve samples. Exact primitives establish reference integrals; curve samples only draw the smooth graph. Controls are temporary and award no mastery.

Why a continuous function attains its average

Let f be continuous on [a,b], with a<b. Compactness supplies an attained minimum m and maximum M. Integral monotonicity gives m(b−a)≤∫f≤M(b−a), so the integral average lies between m and M. The intermediate value theorem gives a point c with f(c) equal to that average. If m=M, every point works. This argument uses the existing extreme and intermediate value theorems and integral properties, rather than defining integrals through a sample point.

Continuity cannot simply be replaced by integrability: the function zero before 1/2 and one from 1/2 onward has integral average 1/2 on [0,1], but takes only zero and one.

Derive the tagged-rectangle error from one common bound

Now suppose f is continuous on [a,b], differentiable inside, and |f′|≤M there. The ordinary mean value theorem gives |f(x)−f(y)|≤M|x−y| for every pair, including endpoints. No continuous derivative is required. For one cell [u,v] and a tag t in it, the rectangle error is the integral of f(t)−f(x). The absolute integral bound gives

|f(t)(v−u)−∫ᵤᵛf(x)dx| ≤ M∫ᵤᵛ|x−t|dx = (M/2)((t−u)²+(v−t)²).

Sum these local bounds using the triangle inequality. Signed local errors may cancel; a guaranteed bound does not assume they will. For right or left endpoints, the total is (M/2)ΣΔ², where Δ is each cell’s width. Since ΣΔ²≤mesh(P)(b−a), refinement with mesh tending to zero guarantees convergence. On [0,1] with n equal cells this gives M/(2n), which also proves the weaker M/n estimate requested in MIT 18.100B problem set 9, question 2.

For midpoint tags the same Lipschitz proof gives (M/4)ΣΔ². This is still a first-derivative bound, not a claim of second-order midpoint accuracy for every Lipschitz function. A sharper order requires additional smoothness or a different argument. The numerical quadrature lesson treats those stronger hypotheses.

For f(x)=2x+1, M=2, endpoint errors achieve the local bound: their signs are uniform and the absolute total equals ΣΔ². Midpoint errors cancel exactly within every cell. For a jump function, a derivative bound away from the jump is insufficient: MVT cannot cross that jump. With two equal cells and right tags, the step gives Q=1 and I=1/2, although its derivative is zero wherever it exists. The inspector withholds the global certificate instead of using that invalid zero bound.

On a graded partition, preserve every width. For x² with boundaries 0, 1/4 and 1, the right sum is 49/64, its exact integral is 1/3, and the M=2 bound is 5/8. Refinement, tags and unequal weights have distinct effects; inspect the local ledger before interpreting the total error.

Continue

Optional depth connects tagged Riemann sums to Darboux bounds. Any tagged sum for a fixed partition lies between its lower and upper sums. To prove the full equivalence with a mesh-based Riemann definition, one must also control arbitrary sufficiently fine partitions, not only a chosen sequence of convenient partitions.

Save a derivation of the refinement inequalities and the small-gap criterion. These are the tools used next to prove integrability of continuous and monotone functions. No antiderivative calculation should appear in those proofs, since the rigorous relationship between integration and differentiation has not yet been established.